Respuesta :
The answer is directrix is horizontal, so the parabola is vertical
focus lies below the directrix, so the parabola opens downwards.
General equation for down-opening parabola:
y = a(x-h)²+k
with
a<0
vertex (h,k)
focal length p = 1/|4a|
focus (h,k-p)
directrix (h,k+p)
Apply your data and solve for h, k, and a.
vertex is halfway between focus and directrix: (3,3)
h = 3
k = 3
distance between focus and vertex is 2, so p = 2.
a = -1/(4p) = -⅛
y = -⅛(x-3)²+3
focus lies below the directrix, so the parabola opens downwards.
General equation for down-opening parabola:
y = a(x-h)²+k
with
a<0
vertex (h,k)
focal length p = 1/|4a|
focus (h,k-p)
directrix (h,k+p)
Apply your data and solve for h, k, and a.
vertex is halfway between focus and directrix: (3,3)
h = 3
k = 3
distance between focus and vertex is 2, so p = 2.
a = -1/(4p) = -⅛
y = -⅛(x-3)²+3
(x-h)^2=4p(y-k)
vertex is (h,k)
p is the distance from focus to vertex and distance from vertex to directix (vertex is in middle of directix and focus)
if p is positive, the parabola opens up and the focus is above the directix
if p is negative, the parabola opens down and the focus is below the directix
we see the directix is over the focus (1>-1) so the parabola opens down and p is negative
distance from (5,-1) to y=1 is 2 units
2/2=1
p=-1 since p is negative
1 up from (5,-1) is (5,0)
veretx is (5,0)
(x-5)^2=4(-1)(y-0)
(x-5)^2=-4y is the equation
vertex is (h,k)
p is the distance from focus to vertex and distance from vertex to directix (vertex is in middle of directix and focus)
if p is positive, the parabola opens up and the focus is above the directix
if p is negative, the parabola opens down and the focus is below the directix
we see the directix is over the focus (1>-1) so the parabola opens down and p is negative
distance from (5,-1) to y=1 is 2 units
2/2=1
p=-1 since p is negative
1 up from (5,-1) is (5,0)
veretx is (5,0)
(x-5)^2=4(-1)(y-0)
(x-5)^2=-4y is the equation